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frez [133]
3 years ago
14

PLEASE HELP IM STUCK ON THIS I STUDIED AND I STILL DONT GET IT I ASKED TEACHERS BUT THEY WONT HELP PLZ HELP

Physics
1 answer:
LenaWriter [7]3 years ago
8 0

3) by waves traveling with or without a medium (also called radiation)

this is radiant energy from the sun; it's the only kind of energy we can see with our eyes


You might be interested in
Refer to the first diagram. What is the weight of the person hanging on the end of the seesaw in Newtons?
irina1246 [14]

Due to equilibrium of moments:

1) The weight of the person hanging on the left is 250 N

2) The 400 N person is 3 m from the fulcrum

3) The weight of the board is 200 N

Explanation:

1)

To solve the problem, we use the principle of equilibrium of moments.

In fact, for the seesaw to be in equilibrium, the total clockwise moment must be equal to the total anticlockwise moment.

The moment of a force is defined as:

M=Fd

where

F is the magnitude of the force

d is the perpendicular distance of the force from the fulcrum

In the first diagram:

- The clockwise moment is due to the person on the right is

M_c = W_2 d_2

where W_2 = 500 N is the weight of the person and d_2 = 2 m is its distance from the fulcrum

- The anticlockwise moment due to the person hanging on the left is

M_a = W_1 d_1

where W_1 is his weight and d_1 = 4 m is the distance from the fulcrum

Since the seesaw is in equilibrium,

M_c = M_a

So we can find the weight of the person on the left:

W_1 d_1 = W_2 d_2\\W_1 = \frac{W_2 d_2}{d_1}=\frac{(500)(2)}{4}=250 N

2)

Again, for the seesaw to be in equilibrium, the total clockwise moment must be equal to the total anticlockwise moment.

- The clockwise moment due to the person on the right is

M_c = W_2 d_2

where W_2 = 400 N is the weight of the person and d_2 is its distance from the fulcrum

- The anticlockwise moment due to the person on the left is

M_a = W_1 d_1

where W_1 = 300 N is his weight and d_1 = 4 m is the distance from the fulcrum.

Since the seesaw is in equilibrium,

M_c = M_a

So we can find the distance of the person on the right:

W_1 d_1 = W_2 d_2\\d_2 = \frac{W_1 d_1}{W_2}=\frac{(300)(4)}{400}=3 m

3)

As before, for the seesaw to be in equilibrium, the total clockwise moment must be equal to the total anticlockwise moment.

- The clockwise moment around the fulcrum this time is due to the weight of the seesaw:

M_c = W_2 d_2

where W_2 is the weight of the seesaw and d_2 = 3 m is the distance of its centre of mass from the fulcrum

- The anticlockwise moment due to the person on the left is

M_a = W_1 d_1

where W_1 = 600 N is his weight and d_1 = 1 m is the distance from the fulcrum

Since the seesaw is in equilibrium,

M_c = M_a

So we can find the weight of the seesaw:

W_1 d_1 = W_2 d_2\\W_2 =\frac{W_1 d_1}{d_2}= \frac{(600)(1)}{3}=200 N

#LearnwithBrainly

8 0
3 years ago
True or False?
olga nikolaevna [1]

Answer:

False

Explanation:

It is better to be logical when proving a point

6 0
2 years ago
A disk of radius a has a total charge Q uniformly distributed over its surface. The disk has negligible thickness and lies in th
sleet_krkn [62]

The electric potential V(z) on the z-axis is :  V = (\frac{Q}{a^2} ) [ (a^2 + z^2)^{\frac{1}{2} } -z

The magnitude of the electric field on the z axis is : E = kб 2\pi( 1 - [z / √(z² + a² ) ] )

<u>Given data :</u>

V(z) =2kQ / a²(v(a² + z²) ) -z  

<h3>Determine the electric potential V(z) on the z axis and magnitude of the electric field</h3>

Considering a disk with radius R

Charge = dq

Also the distance from the edge to the point on the z-axis = √ [R² + z²].

The surface charge density of the disk ( б ) = dq / dA

Small element charge dq =  б( 2πR ) dr

dV  \frac{k.dq}{\sqrt{R^2+z^2} } \\\\= \frac{k(\alpha (2\pi R)dR}{\sqrt{R^2+z^2} }  ----- ( 1 )

Integrating equation ( 1 ) over for full radius of a

∫dv = \int\limits^a_o {\frac{k(\alpha (2\pi R)dR)}{\sqrt{R^2+z^2} } } \,

 V = \pi k\alpha [ (a^2+z^2)^\frac{1}{2} -z ]

     = \pi k (\frac{Q}{\pi \alpha ^2})[(a^2 +z^2)^{\frac{1}{2} }  -z ]

Therefore the electric potential V(z) = (\frac{Q}{a^2} ) [ (a^2 + z^2)^{\frac{1}{2} } -z

Also

The magnitude of the electric field on the z axis is : E = kб 2\pi( 1 - [z / √(z² + a² ) ] )

Hence we can conclude that the answers to your question are as listed above.

Learn more about electric potential : brainly.com/question/25923373

7 0
2 years ago
A student jobs around a square park two times. Starting and ending at the gate to the park. The square jogging track is 40 meter
olga nikolaevna [1]
The answer would be the first one
3 0
3 years ago
Read 2 more answers
Calculate the work efecttuated, when the force is 300 N, and the distance is 300 meters.
Alexxx [7]

Hello!

Use the formula:

W = Fd

Replacing:

W = 300 N * 300 m

Resolving:

W = 90000 J

The work is <u>90000 Joules.</u>

8 0
3 years ago
Read 2 more answers
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